习题课 ■解: du. ·延长OC至D,过C做CB∥Ox轴,则有∠DCB=0: ·又因AN=M, 故有∠MCO=0: ·从而∠BCMM=π-20.0 ·则矢量CM与x轴正向的交角为:-(π-20) ·即CM=acos[-(t-20)]i+asin[-(π-20)]j ·于是有OC=2acos0i+2asin0j ·从而F=O应M=0C+CM =(2acose-acos20)i+(2asin0-asin20)j lexu@mail.xidian.edu.cnlexu@mail.xidian.edu.cn